Question

paired t-test
goodness-of-fit
b. what are the null and alternative hypotheses?
$h_0$ :
the distribution of hair color is the same for each body type.
the distribution of hair color is not the same for each body type.
hair color and body type are dependent.
hair color and body type are independent.
$h_1$ :
the distribution of hair color is the same for each body type.
hair color and body type are dependent.
the distribution of hair color is not the same for each body type.
hair color and body type are independent.
c. the test-statistic for this data = (please show your answer to three decimal places.)
d. the p-value for this sample = (please show your answer to four decimal places.)

Explanation:

Step1: Recall Chi-Square Test for Independence

For a chi - square test of independence (which is related to goodness - of - fit in the context of two categorical variables), the test statistic formula is $\chi^{2}=\sum\frac{(O - E)^{2}}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency. But since the problem does not provide the observed and expected frequencies, we assume that maybe this is a follow - up question where the data (observed and expected frequencies) was given earlier. Let's assume we have a contingency table with hair color and body type as the two variables.

Step2: Calculate Expected Frequencies (if not given)

Suppose we have a contingency table with rows representing hair color and columns representing body type. Let $r_{i}$ be the row totals, $c_{j}$ be the column totals, and $N$ be the grand total. The expected frequency $E_{ij}=\frac{r_{i}\times c_{j}}{N}$.

Step3: Calculate the Test Statistic

Once we have the observed frequencies $O_{ij}$ and expected frequencies $E_{ij}$, we calculate $\chi^{2}=\sum_{i,j}\frac{(O_{ij}-E_{ij})^{2}}{E_{ij}}$. After performing the calculations (assuming the data was provided in the original problem context), we round the result to three decimal places.

(Note: Since the actual data is not provided in the current question, but if we assume a typical example, for instance, if we had a contingency table like:

| Hair Color\Body Type | Type 1 | Type 2 | Type 3 | Row Totals |
|---|---|---|---|---|
| Blonde | 20 | 15 | 10 | 45 |
| Brunette | 30 | 25 | 20 | 75 |
| Red | 10 | 10 | 5 | 25 |
| Column Totals | 60 | 50 | 35 | 145 |

First, calculate expected frequencies:

$E_{11}=\frac{45\times60}{145}\approx18.6207$

$E_{12}=\frac{45\times50}{145}\approx15.5172$

$E_{13}=\frac{45\times35}{145}\approx10.8621$

$E_{21}=\frac{75\times60}{145}\approx30.1034$

$E_{22}=\frac{75\times50}{145}\approx25.8621$

$E_{23}=\frac{75\times35}{145}\approx18.1034$

$E_{31}=\frac{25\times60}{145}\approx10.3448$

$E_{32}=\frac{25\times50}{145}\approx8.6207$

$E_{33}=\frac{25\times35}{145}\approx6.0345$

Then calculate $(O - E)^{2}/E$ for each cell:

For (Blonde, Type 1): $\frac{(20 - 18.6207)^{2}}{18.6207}\approx\frac{(1.3793)^{2}}{18.6207}\approx\frac{1.9024}{18.6207}\approx0.1022$

For (Blonde, Type 2): $\frac{(15 - 15.5172)^{2}}{15.5172}\approx\frac{(- 0.5172)^{2}}{15.5172}\approx\frac{0.2675}{15.5172}\approx0.0172$

For (Blonde, Type 3): $\frac{(10 - 10.8621)^{2}}{10.8621}\approx\frac{(-0.8621)^{2}}{10.8621}\approx\frac{0.7432}{10.8621}\approx0.0684$

For (Brunette, Type 1): $\frac{(30 - 30.1034)^{2}}{30.1034}\approx\frac{(-0.1034)^{2}}{30.1034}\approx\frac{0.0107}{30.1034}\approx0.0004$

For (Brunette, Type 2): $\frac{(25 - 25.8621)^{2}}{25.8621}\approx\frac{(-0.8621)^{2}}{25.8621}\approx\frac{0.7432}{25.8621}\approx0.0287$

For (Brunette, Type 3): $\frac{(20 - 18.1034)^{2}}{18.1034}\approx\frac{(1.8966)^{2}}{18.1034}\approx\frac{3.5971}{18.1034}\approx0.1987$

For (Red, Type 1): $\frac{(10 - 10.3448)^{2}}{10.3448}\approx\frac{(-0.3448)^{2}}{10.3448}\approx\frac{0.1189}{10.3448}\approx0.0115$

For (Red, Type 2): $\frac{(10 - 8.6207)^{2}}{8.6207}\approx\frac{(1.3793)^{2}}{8.6207}\approx\frac{1.9024}{8.6207}\approx0.2207$

For (Red, Type 3): $\frac{(5 - 6.0345)^{2}}{6.0345}\approx\frac{(-1.0345)^{2}}{6.0345}\approx\frac{1.0702}{6.0345}\approx0.1773$

Now sum all these values:

$0.1022 + 0.0172+0.0684 + 0.0004+0.0287+0.1987+0.0115+0.2207+0.1773\approx0.8251$

Rounding to three decimal places, we get $0.825$. But this is just an example. The actual value depends on the given data.)

Answer:

(The answer will be the calculated test statistic rounded to three decimal places. If we assume the above example, the answer is 0.825. But in the actual problem, you need to use the given observed and expected frequencies to calculate it.)